\beginsection 18.1

Let $f$ be as in Theorem 18.1. Show that if $-f$ assumes its maximum at
$x_0\in[a,b]$, then $f$ assumes its minimum at $x_0$.

\medskip
Since $x_0$ is a maximum for $-f$ we have
$$-f(x_0)\ge-f(x).$$
Multiply through by $-1$ to get
$$f(x_0)\le f(x).$$
We observe that $x_0$ is a minimum for $f$.

